Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.
Table: Users
+---------------+---------+ | Column Name | Type | +---------------+---------+ | id | int | | name | varchar | +---------------+---------+ id is the column with unique values for this table. name is the name of the user.
Table: Rides
+---------------+---------+ | Column Name | Type | +---------------+---------+ | id | int | | user_id | int | | distance | int | +---------------+---------+ id is the column with unique values for this table. user_id is the id of the user who traveled the distance "distance".
Write a solution to report the distance traveled by each user.
Return the result table ordered by travelled_distance in descending order, if two or more users traveled the same distance, order them by their name in ascending order.
The result format is in the following example.
Example 1:
Input: Users table: +------+-----------+ | id | name | +------+-----------+ | 1 | Alice | | 2 | Bob | | 3 | Alex | | 4 | Donald | | 7 | Lee | | 13 | Jonathan | | 19 | Elvis | +------+-----------+ Rides table: +------+----------+----------+ | id | user_id | distance | +------+----------+----------+ | 1 | 1 | 120 | | 2 | 2 | 317 | | 3 | 3 | 222 | | 4 | 7 | 100 | | 5 | 13 | 312 | | 6 | 19 | 50 | | 7 | 7 | 120 | | 8 | 19 | 400 | | 9 | 7 | 230 | +------+----------+----------+ Output: +----------+--------------------+ | name | travelled_distance | +----------+--------------------+ | Elvis | 450 | | Lee | 450 | | Bob | 317 | | Jonathan | 312 | | Alex | 222 | | Alice | 120 | | Donald | 0 | +----------+--------------------+ Explanation: Elvis and Lee traveled 450 miles, Elvis is the top traveler as his name is alphabetically smaller than Lee. Bob, Jonathan, Alex, and Alice have only one ride and we just order them by the total distances of the ride. Donald did not have any rides, the distance traveled by him is 0.
Problem summary: Table: Users +---------------+---------+ | Column Name | Type | +---------------+---------+ | id | int | | name | varchar | +---------------+---------+ id is the column with unique values for this table. name is the name of the user. Table: Rides +---------------+---------+ | Column Name | Type | +---------------+---------+ | id | int | | user_id | int | | distance | int | +---------------+---------+ id is the column with unique values for this table. user_id is the id of the user who traveled the distance "distance". Write a solution to report the distance traveled by each user. Return the result table ordered by travelled_distance in descending order, if two or more users traveled the same distance, order them by their name in ascending order. The result format is in the following example.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers":{"Users":["id","name"],"Rides":["id","user_id","distance"]},"rows":{"Users":[[1,"Alice"],[2,"Bob"],[3,"Alex"],[4,"Donald"],[7,"Lee"],[13,"Jonathan"],[19,"Elvis"]],"Rides":[[1,1,120],[2,2,317],[3,3,222],[4,7,100],[5,13,312],[6,19,50],[7,7,120],[8,19,400],[9,7,230]]}}Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1407: Top Travellers
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #1407: Top Travellers
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1407: Top Travellers
// # Write your MySQL query statement below
// SELECT name, IFNULL(SUM(distance), 0) AS travelled_distance
// FROM
// Users AS u
// LEFT JOIN Rides AS r ON u.id = r.user_id
// GROUP BY u.id
// ORDER BY 2 DESC, 1;
// "#
// }
// Accepted solution for LeetCode #1407: Top Travellers
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1407: Top Travellers
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1407: Top Travellers
// # Write your MySQL query statement below
// SELECT name, IFNULL(SUM(distance), 0) AS travelled_distance
// FROM
// Users AS u
// LEFT JOIN Rides AS r ON u.id = r.user_id
// GROUP BY u.id
// ORDER BY 2 DESC, 1;
// "#
// }
# Accepted solution for LeetCode #1407: Top Travellers
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #1407: Top Travellers
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #1407: Top Travellers
# # Write your MySQL query statement below
# SELECT name, IFNULL(SUM(distance), 0) AS travelled_distance
# FROM
# Users AS u
# LEFT JOIN Rides AS r ON u.id = r.user_id
# GROUP BY u.id
# ORDER BY 2 DESC, 1;
# "#
# }
// Accepted solution for LeetCode #1407: Top Travellers
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #1407: Top Travellers
# Write your MySQL query statement below
SELECT name, IFNULL(SUM(distance), 0) AS travelled_distance
FROM
Users AS u
LEFT JOIN Rides AS r ON u.id = r.user_id
GROUP BY u.id
ORDER BY 2 DESC, 1;
"#
}
// Accepted solution for LeetCode #1407: Top Travellers
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1407: Top Travellers
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1407: Top Travellers
// # Write your MySQL query statement below
// SELECT name, IFNULL(SUM(distance), 0) AS travelled_distance
// FROM
// Users AS u
// LEFT JOIN Rides AS r ON u.id = r.user_id
// GROUP BY u.id
// ORDER BY 2 DESC, 1;
// "#
// }
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.